@^ he has already asked to optimize n^2logn solution

On Wed, Jun 16, 2010 at 5:25 PM, divya jain <[email protected]>wrote:

> sort array c only.
> now select every pair from array a nd b ( o(n2))
> for every pair find the element from c which gives sum of the three element
> 0. ( search using binary search as c is sorted)
> so the algo is O (n^2 logn))
>
>
> On 16 June 2010 13:47, Amir hossein Shahriari <
> [email protected]> wrote:
>
>> @algoose chase: you're right thx for the test case
>>
>> On Wed, Jun 16, 2010 at 11:45 AM, Algoose Chase <[email protected]>wrote:
>>
>>> @ Amir:
>>> I think you cannot find two numbers in B and C such that their sum is
>>> -A[k] in O(n) in all cases with the logic that you mentioned.
>>>
>>> for instance you can try with :
>>> A : 2 7 8 10
>>> B :-2, 0, 1, 9
>>> C: -7 -2 1 3
>>>
>>> On Wed, Jun 16, 2010 at 5:56 AM, Amir hossein Shahriari <
>>> [email protected]> wrote:
>>>
>>>> sort all arrays
>>>> for each number in A , A[k]
>>>> find two numbers in B and C such that their sum is -A[k]
>>>> this can be done in O(n) time:
>>>>
>>>> set i at the beginning of B and j at the end of C
>>>> while i<n or j>=0
>>>>   if ( B[i] + C[j] == -A[k] ) return true
>>>>   if ( B[i] + C[j] < -A[k] ) increase i
>>>>   else decrease j
>>>>
>>>> we have to do this for each element of A so the overall complexity is:
>>>> O(n2) time O(1) space
>>>>
>>>> correct me if I'm wrong
>>>>
>>>> On 6/15/10, sharad kumar <[email protected]> wrote:
>>>> > plzzz comment on this question
>>>> >
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Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD

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